Multivariable Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 0-53849-787-4
ISBN 13: 978-0-53849-787-9

Chapter 16 - Vector Calculus - Review - Exercises - Page 1162: 33

Answer

$-\dfrac{1}{2}$

Work Step by Step

Apply Stoke's Theorem $\iint_S curl F\cdot dS=\int_C F \cdot dr$ The equation of the plane with the intercepts can be written as: $\dfrac{x}{1}+\dfrac{y}{1}+\dfrac{z}{1}=1$ $\implies z=1-x-y$ We know that curl $F=\nabla \times F$ Here, we have curl $F=-yi-zj-xk$ $\iint_S curl F\cdot dS=\iint_D[-y-z-x] dA=-\iint_D dA$ Here, $\iint_D dA$ shows the area of the triangle, which is equal to $=\dfrac{1}{2}(1)(1)=\dfrac{1}{2}$ Therefore, $\iint_S curl F\cdot dS=\int_C F \cdot dr=-\iint_D dA=-\dfrac{1}{2}$
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